What is the solution for x^{''}+4x^'+13x=0 ?

The solution for x^{''}+4x^'+13x=0 is x=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))

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x^{''}+4x^'+13x=0

x^{^{''}} + 4 x^{^{'}} + 13 x = 0

x = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

x^{''} (t) + 4 x^{'} (t) + 13 x = 0

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

x = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

What is the solution for x^{''}+4x^'+13x=0 ?
The solution for x^{''}+4x^'+13x=0 is x=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))